Ice at 0 degree celcuis is added to 200g of water initially at 70 degree celcuis in a vacuum flask.When 50g of ice has been added and has all melted the temperature of the flask and contents is 40 degree Celsius when a further 80gm of ice has been added and has all melted the temperature of the whole is 10 degree Celsius calculate the specific latent heat of the
fusion of ice.
First, let's consider the energy transfers when the first 50g of ice are added.
The energy gained by the ice when it melts is given by: Q1 = m1 * L, where m1 is the mass of the ice (50g) and L is the specific latent heat of fusion (which we are trying to find).
The energy lost by the 200g of water when the temperature drops from 70°C to 40°C is given by: Q2 = m2 * c * ΔT2, where m2 is the mass of the water (200g), c is the specific heat capacity of water (4.18 J/g°C), and ΔT2 is the change in temperature (70 - 40 = 30°C).
Since the system is thermally isolated, the energy gained by the ice should be equal to the energy lost by the water. Thus, Q1 = Q2.
50 * L = 200 * 4.18 * 30
L = (200 * 4.18 * 30) / 50
L = 502.8 J/g
Now, let's consider the energy transfers when an additional 80g of ice is added.
The energy gained by the ice when it melts is given by: Q3 = m3 * L, where m3 is the mass of the ice (80g) and L is the specific latent heat of fusion (which we found earlier to be 502.8 J/g).
The energy lost by the 250g of water and melted ice when the temperature drops from 40°C to 10°C is given by: Q4 = m4 * c * ΔT4, where m4 is the mass of the water (200g + 50g = 250g), c is the specific heat capacity of water (4.18 J/g°C), and ΔT4 is the change in temperature (40 - 10 = 30°C).
Again, since the system is thermally isolated, the energy gained by the ice should be equal to the energy lost by the water. Thus, Q3 = Q4.
80 * L = 250 * 4.18 * 30
L = (250 * 4.18 * 30) / 80
L = 393.6 J/g
Since we got different values for the specific latent heat of fusion in each case, we can take the average to get a more accurate value.
L = (502.8 + 393.6) / 2
L = 448.2 J/g
The specific latent heat of fusion of ice is approximately 448.2 J/g.
To calculate the specific latent heat of the substance used, we can use the formula:
Q = mL
where Q is the heat energy transferred, m is the mass of the substance, and L is the specific latent heat of the substance.
Let's break down the problem step by step:
Step 1: Calculate the heat energy transferred when 50g of ice melts.
We can use the formula:
Q = mL
where Q is the heat energy transferred, m is the mass of the substance, and L is the specific latent heat of the substance.
Given:
m = 50g (mass of ice)
L = ? (specific latent heat of the substance)
Since the ice melts at 0°C, we need to determine the heat energy needed to raise the temperature of ice from 0°C to the final temperature of the water:
Q = m * (final temperature - initial temperature) * specific heat capacity
The final temperature is given as 40°C, and the initial temperature is 0°C.
Assuming the specific heat capacity of ice is 2.09 J/g°C, we can calculate the heat energy transferred:
Q = 50g * (40°C - 0°C) * 2.09 J/g°C
Q = 4170 J
Step 2: Calculate the specific latent heat of the substance.
Since the ice has completely melted, the heat energy transferred is equal to the latent heat:
Q = mL
Substituting the known values:
4170 J = 50g * L
L = 4170 J / 50g
L = 83.4 J/g
Therefore, the specific latent heat of the substance is 83.4 J/g.
To solve this problem, we need to apply the principle of conservation of energy. The energy gained by the ice as it melts is equal to the energy lost by the water as it cools down. We can use the equation:
Q = m × L
where:
Q is the energy transferred,
m is the mass of the substance,
L is the specific latent heat of the substance.
Given data:
Initial mass of water, m1 = 200g
Initial temperature of water, T1 = 70°C
Mass of ice added after the first melting, m2 = 50g
Final temperature after the first melting, T2 = 40°C
Mass of ice added after the second melting, m3 = 80g
Final temperature after the second melting, T3 = 10°C
Let's calculate the energy transferred in each step:
1. The energy transferred between the water and ice during the first melting:
Energy gained by ice = m2 × L
Energy lost by water = m1 × c × (T1 - T2)
where c is the specific heat capacity of water.
2. The energy transferred between the water and ice during the second melting:
Energy gained by ice = m3 × L
Energy lost by water = (m1 + m2) × c × (T2 - T3)
Now we can set up the equations and solve them step by step:
1. Calculate the energy transferred during the first melting:
m2 × L = m1 × c × (T1 - T2)
2. Calculate the energy transferred during the second melting:
m3 × L = (m1 + m2) × c × (T2 - T3)
3. Rearrange the equations to solve for L:
L = (m1 × c × (T1 - T2)) / m2
L = ((m1 + m2) × c × (T2 - T3)) / m3
4. Calculate the average value of L:
Average L = (L1 + L2) / 2
where L1 is the value of L calculated in step 3 for the first melting, and L2 is the value of L calculated for the second melting.
By substituting the values into these equations, you can determine the specific latent heat of the substance.